Chapter 3.11, Problem 169E

This exercise demonstrates that, in general, the results provided by Tchebysheff’s theorem cannot be improved upon. Let Y be a random variable such that

$p\left(-1\right)=\frac{1}{18},p\left(0\right)=\frac{16}{18},p\left(1\right)=\frac{1}{18}.$

a Show that E(Y) = 0 and V(Y) = 1/9.

b Use the probability distribution of Y to calculate $P\left(\left|Y-\mu \right|\ge 3\sigma \right)$. Compare this exact probability with the upper bound provided by Tchebysheff’s theorem to see that the bound provided by Tchebysheff’s theorem is actually attained when k = 3.

*c In part (b) we guaranteed E(Y) = 0 by placing all probability mass on the values –1, 0, and 1, with p(–1) = p(1). The variance was controlled by the probabilities assigned to p(–1) and p(1). Using this same basic idea, construct a probability distribution for a random variable X that will yield $P\left(\left|X-{\mu }_X\right|\ge 2{\sigma }_X\right)=1/4$.

*d If any k > 1 is specified, how can a random variable W be constructed so that $P\left(\left|W-{\mu }_W\right|\ge k{\sigma }_W\right)=1/k^2$?

To determine

Prove that $E\left(Y\right)=0\text{and }V\left(Y\right)=\frac{1}{9}$.

Explanation of Solution

Calculation:

The expected value of a random variable is the sum of the product of each observation with corresponding probability.

Hence, the expected value is obtained as follows:

$\begin{array}{c}E\left(Y\right)={\displaystyle \sum _{y}yp\left(y\right)}\\ =\left(-1\right)\left(\frac{1}{18}\right)+\left(0\right)\left(\frac{16}{18}\right)+\left(1\right)\left(\frac{1}{18}\right)\\ =0\end{array}$

Similarly the variance of a random variable is the expected value of the squared deviation from the mean.

$\begin{array}{c}V\left(Y\right)={\displaystyle \sum _y{\left(y-E\left(Y\right)\right)}^{2}p\left(y\right)}\\ ={\left(-1-0\right)}^2\left(\frac{1}{18}\right)+{\left(0-0\right)}^2\left(\frac{16}{18}\right)+{\left(1-0\right)}^2\left(\frac{1}{18}\right)\\ =\frac{1}{9}\end{array}$

Hence, it is proved that $E\left(Y\right)=0\text{and }V\left(Y\right)=\frac{1}{9}$.

To determine

Find the value of $P\left(\left|Y-\mu \right|\ge 3\sigma \right)$.

Compare the exact probability with the upper bound of the Tchebysheff’s theorem.

Answer to Problem 169E

The value of $P\left(\left|Y-\mu \right|\ge 3\sigma \right)$ is $\frac{1}{9}$.

Explanation of Solution

Calculation:

Tchebysheff’s Theorem:

Let Y be a random variable with mean $\mu$ and finite variance ${\sigma }^2$. Then, for any constant $k>0$ it can be said that,

$P\left(\left|Y-\mu \right|<k\sigma \right)\ge 1-\frac{1}{k^2}$ or $P\left(\left|Y-\mu \right|\ge k\sigma \right)\le \frac{1}{k^2}$.

Thus, using the Tchebysheff’s theorem it can be obtained that,

$\begin{array}{c}P\left(\left|Y-\mu \right|<k\sigma \right)\ge 1-\frac{1}{k^2}\\ P\left(\left|Y-\mu \right|\ge k\sigma \right)\ge \frac{1}{k^2}\\ P\left(\left|Y-\mu \right|\ge 3\sigma \right)=\frac{1}{3^2}\\ =\frac{1}{9}\end{array}$

Similarly, using the given information the required probability is obtained as follows:

$\begin{array}{c}P\left(\left|Y-\mu \right|\ge 3\sigma \right)=P\left(\left|Y-0\right|\ge 3\left(\frac{1}{3}\right)\right)\\ =P\left(\left|Y\right|\ge 1\right)\\ =P\left(Y=-1\right)+P\left(Y=1\right)\\ =\frac{1}{18}+\frac{1}{18}\\ =\frac{1}{9}\end{array}$

Hence, the value of $P\left(\left|Y-\mu \right|\ge 3\sigma \right)$ is $\frac{1}{9}$.

Thus, it can be said the exact probability is same with the upper bound of the Tchebysheff’s theorem when $k=3$ .

To determine

Generate a probability distribution for a random variable X that will yield $P\left(\left|X-{\mu }_X\right|\ge 2{\sigma }_X\right)=\frac{1}{4}$

Answer to Problem 169E

The probability distribution of another sufficient random variable is,

 $p\left(x\right)=\{\begin{array}{c}\frac{1}{8},x=-1\\ \frac{3}{4},x=0\\ \frac{1}{8},x=1\end{array}$

Explanation of Solution

Calculation:

Consider the random variable X which is symmetric around mean 0.

The probability distribution of X is,

$p\left(x\right)=\{\begin{array}{l}p,x=-1\\ 0,x=0\\ p,x=1\end{array}$.

It is needed to find the value of p.

The mean of X is,

$\begin{array}{c}E\left(X\right)=\left(1\right)\left(p\right)+0+\left(-1\right)\left(p\right)\\ =0\end{array}$

The variance of X is,

$\begin{array}{c}V\left(X\right)={\left(1-0\right)}^2\left(p\right)+0+{\left(-1-0\right)}^2\left(p\right)\\ =2p\end{array}$

Thus, using Tchebysheff’s theorem it can be said that,

$\begin{array}{c}P\left(\left|X-{\mu }_X\right|>2{\sigma }_X\right)=P\left(\left|X-0\right|>2\sqrt{2p}\right)\\ =2P\left(X>2\sqrt{2p}\right)\end{array}$

Now, it is obvious that $2P\left(X>2\sqrt{2p}\right)=\frac{1}{4}$ if and only if $2\sqrt{2p}\le 1$.

Hence,

$\begin{array}{c}2\sqrt{2p}\le 1\\ p\le \frac{1}{8}\end{array}$ .

Thus, the probability distribution of another sufficient random variable is,

 $\begin{array}{l}p\left(x\right)=\{\begin{array}{l}\frac{1}{8},x=-1\\ 1-\left(\frac{1}{8}+\frac{1}{8}\right),x=0\\ \frac{1}{8},x=1\end{array}\\ =\{\begin{array}{c}\frac{1}{8},x=-1\\ \frac{3}{4},x=0\\ \frac{1}{8},x=1\end{array}\end{array}$

To determine

Explain the procedure of obtaining a random variable W such that $P\left(\left|W-{\mu }_w\right|\ge k{\sigma }_w\right)=\frac{1}{k^2}$  if any $k>1$ is specified.

Explanation of Solution

Calculation:

Consider the random variable W which is symmetric around mean 0.

The probability distribution of W is,

$p\left(w\right)=\{\begin{array}{l}p,w=-1\\ 0,w=0\\ p,w=1\end{array}$.

It is needed to find the value of p.

The mean of W is,

$\begin{array}{c}E\left(W\right)=\left(1\right)\left(p\right)+0+\left(-1\right)\left(p\right)\\ =0\end{array}$

The variance of W is,

$\begin{array}{c}V\left(W\right)={\left(1-0\right)}^2\left(p\right)+0+{\left(-1-0\right)}^2\left(p\right)\\ =2p\end{array}$

Thus, using Tchebysheff’s theorem it can be said that,

$\begin{array}{c}P\left(\left|W-{\mu }_w\right|\ge k{\sigma }_w\right)=\frac{1}{k^2}\\ P\left(\left|W-0\right|>k\sqrt{2p}\right)=\frac{1}{k^2}\\ 2P\left(W>k\sqrt{2p}\right)=\frac{1}{k^2}\end{array}$

It is clear that W can take only one positive value. Hence, if $k\sqrt{2p}\le 1$, then only it is possible that $W>k\sqrt{2p}$.

Hence,

$\begin{array}{c}k\sqrt{2p}\le 1\\ p\le \frac{1}{2k^2}\end{array}$ .

Thus, the probability distribution of another sufficient random variable is,

 $\begin{array}{l}p\left(w\right)=\{\begin{array}{l}\frac{1}{2k^2},w=-1\\ 1-\left(\frac{2}{2k^2}\right),w=0\\ \frac{1}{2k^2},w=1\end{array}\\ =\{\begin{array}{l}\frac{1}{2k^2},w=-1\\ 1-\left(\frac{1}{k^2}\right),w=0\\ \frac{1}{2k^2},w=1\end{array}\end{array}$